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3 years ago

SOMEONE HELPP MEEE PLEASEEE THIS IS DUE IN 15 MINS

Mathematics

1 answer:

IrinaK [193]3 years ago

4 0

Answer:

30=3x+6

Step-by-step explanation:

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Find the surface area of the prism.

castortr0y [4]

Answer:

30 i think

Step-by-step explanation:

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3 years ago

The Boardman family walked two over five of a mile in two over seven of an hour. What is their unit rate in miles per hour?

hichkok12 [17]

If the family walked 2/5 of a mile in 2/7 of an hour their MPH would be 2.8mph

3 0

3 years ago

Ginny factored 6x2 – 31x – 30 as shown: ac = –180 and b = 31 36(–5) = –180 and 36 + (–5) = 31 6x2 + 36x – 5x – 30 6x(x + 6) – 5(

meriva

Answer:

Ginny made a mistake in step 1 when she identified b = 31. It should be b=-31

Step-by-step explanation:

Ginny factored $6x^2 -31x -30$ as shown:

ac = –180 and

b = 31

36(–5) = –180 and 36 + (–5) = 31

$6x^2$ + 36x – 5x – 30

6x(x + 6) – 5(x + 6)

(x + 6)(6x – 5)

b is the coefficient of x and from the trinomial $6x^2 -31x -30$ , b=-31.

Therefore Ginny made a mistake in step 1 when she identified b = 31. It should be b = –31.

The correct steps are: $6x^2 -31x -30$ :

ac = –180 and

b = -31

36(–5) = –180 and -36+5 = -31

$6x^2-36x +5x -30$

$6x(x-6)+5(x-6)$

$(x-6)(6x+5)$

7 0

3 years ago

How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$

Since we've shown that the left hand side transforms into the right hand side, this verifies the equation is an identity.

4 0

3 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$

6 0

3 years ago